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10 months ago

Central University of Jammu

What is the cut off of Central University of Jammu?

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Nishtha Kaur

Contributor-Level 7

Central University of Jammu accepts CUET PG entrance exams for admission to the postgraduate courses. The University has released the Round 1 seat allotments results for the candidates belonging to All India quotas. Considering the Central University of Jammu cutoff 2025, the score ranged between 15 and 221 for the General All India category candidates. Moreover, the CU Jammu cutoff 2025 for OBC AI category ranged between 14 and 197.

Candidates can refer to the table given below to check out the Central University of Jammu cutoff 2025 for the General AI quota:

Course	2025
M.Sc. in Physics	95
Master of Education (M.Ed.)	140
M.Sc. in Bioinformatics	53
M.Sc. in Biotechnology	100
M.Sc. in Chemistry	82
M.Sc. in Computer Science	45
M.A. in Economics	30
M.Sc. in Environmental Sciences	45
M.A. in History	102
Master of Journalism and Mass Communication (MJMC)	105
Master of Law (LL.M.)	156
M.Sc. in Life Science	70
M.A. in Psychology	221
M.Pharm. in Pharmacuetics	112
M.Pharm. in Pharmacology	117
M.A. in English	70
M.A in Sociology	87
M.A. in Political Science & International Relations	130
M.Sc. in Geology	130
M.Sc. in Statistics	15
M.Sc. in Geography	73
M.A. in Geograpgy	73
M.Sc. in Artificial Intelligence	35
M.Sc. in Data Science and Applied Statistics	50
M.Sc. in Psychology	221
M.A. in Statistics	15

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10 months ago

physics ncert solutions class 11th Oscillations

A simple pendulum of time period 1s and length l is hung from a fixed support at O, such that the bob is at a distance H vertically above A on the ground (Fig) The amplitude is θ o . The string snaps at θ = θ₀ /2. Find the time taken by the bob to hit the ground. Also find distance from A where bob hits the ground. Assume θ₀ to be small so that sin _o ₀ and cos ₀

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Let us assume t=0 when $θ = θ$ _o then $θ$ = $θ$ ₀coswt

Given a seconds pendulum w=2 $π$

$θ$ = $θ_{o} c o s 2 π t$

At time t, let $θ = θ_{o} / 2$

Cos2 $π t_{1} = \frac{1}{2}$ , t₁=1/6

$\frac{d θ}{d t}$ =-( $θ_{o} 2 π$ )sin2 $π t$

t=t₁=1/6

d $\frac{θ}{d t}$ =- $θ_{o} 2 π$ sin $\frac{2 π}{6} = - \sqrt[]{3} π θ_{o}$

the linear velocity is u=- $\sqrt[]{3} π θ_{o} l$

the vertical component is U_y= $- \sqrt[]{3} π l s i n (\frac{θ_{o}}{2})$

the horizontal component U_x=- $- \sqrt[]{3} π l c o s (\frac{θ_{o}}{2})$

at the time it snaps the vertical height is

H’=H+l(1-cos $(\frac{θ_{o}}{2})$ )

Let the time require for fall be t , then

H’= H+l (1 $- c o s (\frac{θ_{o}}{2})$ )

Let the time required for fall be at t then

H’=u_yt+1/2 gt²

1/2gt²+ $\sqrt[]{3} π θ_{o} l s i n \frac{θ_{o}}{2} t - H^{'} = 0$

t= $\frac{- \sqrt[]{3} π θ_{0} s i n \frac{θ_{o}}{2} \pm \sqrt[]{3 π^{2} θ_{o}^{2} {s i n}^{2} \frac{θ_{o}}{2} + 2 g H'}}{g}$

given that $θ_{o}$ is small , hence neglecting terms of order $θ_{o}^{2}$ and higher

t= $\sqrt[]{\frac{2 H^{'}}{g}}$

H’ $\approx H + l (1 - 1)$

t= $\sqrt[]{\frac{2 H}{g}}$

the distance travelled in the x direction is

...more

This is a long answer type question as classified in NCERT Exemplar

Let us assume t=0 when $θ = θ$ _o then $θ$ = $θ$ ₀coswt

Given a seconds pendulum w=2 $π$

$θ$ = $θ_{o} c o s 2 π t$

At time t, let $θ = θ_{o} / 2$

Cos2 $π t_{1} = \frac{1}{2}$ , t₁=1/6

$\frac{d θ}{d t}$ =-( $θ_{o} 2 π$ )sin2 $π t$

t=t₁=1/6

d $\frac{θ}{d t}$ =- $θ_{o} 2 π$ sin $\frac{2 π}{6} = - \sqrt[]{3} π θ_{o}$

the linear velocity is u=- $\sqrt[]{3} π θ_{o} l$

the vertical component is U_y= $- \sqrt[]{3} π l s i n (\frac{θ_{o}}{2})$

the horizontal component U_x=- $- \sqrt[]{3} π l c o s (\frac{θ_{o}}{2})$

at the time it snaps the vertical height is

H’=H+l(1-cos $(\frac{θ_{o}}{2})$ )

Let the time require for fall be t , then

H’= H+l (1 $- c o s (\frac{θ_{o}}{2})$ )

Let the time required for fall be at t then

H’=u_yt+1/2 gt²

1/2gt²+ $\sqrt[]{3} π θ_{o} l s i n \frac{θ_{o}}{2} t - H^{'} = 0$

t= $\frac{- \sqrt[]{3} π θ_{0} s i n \frac{θ_{o}}{2} \pm \sqrt[]{3 π^{2} θ_{o}^{2} {s i n}^{2} \frac{θ_{o}}{2} + 2 g H'}}{g}$

given that $θ_{o}$ is small , hence neglecting terms of order $θ_{o}^{2}$ and higher

t= $\sqrt[]{\frac{2 H^{'}}{g}}$

H’ $\approx H + l (1 - 1)$

t= $\sqrt[]{\frac{2 H}{g}}$

the distance travelled in the x direction is u_xt to the left of where the bob is snapped

X= Uxt= $\sqrt[]{3} π θ_{o} l c o s (\frac{θ_{o}}{2}) \sqrt[]{\frac{2 H}{g}} s$

cos $\frac{θ_{o}}{2} \approx 1$

X= $\sqrt[]{3} π θ_{o} l \sqrt[]{\frac{2 h}{g}} = \sqrt[]{\frac{6 H}{g}} θ_{o} l π$

At the time of snapping , the bob was at a horizontal distance of $l s i n \frac{θ_{o}}{2} \approx l \frac{θ_{0}}{2}$ from A

Thus, the distance of bob from A where it meets the ground is

= $\frac{l θ_{o}}{2} - X = \frac{l θ_{o}}{2} - \sqrt[]{\frac{6 H}{g}} θ_{o} l π$

= $θ_{o} l (\frac{1}{2} - π \sqrt[]{\frac{6 H}{g}} θ_{o} l π)$

= $θ_{o} l (\frac{1}{2} - π \sqrt[]{\frac{6 H}{g}})$

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10 months ago

Stanley Medical College Medicine

My OBC NCL rank is 9115. Can I get Stanley Medical College or Coimbatore Medical College in AIQ or SQ?

0 Follower 4 Views

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10 months ago

physics ncert solutions class 11th Oscillations

A tunnel is dug through the centre of the Earth. Show that a body of mass ‘m’ when dropped from rest from one end of the tunnel will execute simple harmonic motion.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

The gravitational force on the particle at a distance r from the centre of the earth arises entirely from that portion of matter of the earth in shells internal to the positiin of the particles . the external shells exert no force on the particle.

Let g’ be the acceleration at P

So g’ =g (1-d/R)=g (R-d/R)

R-d=y

g’=gy/R’

F=-mg’= -mgy/R

F $\propto - y$

Ma=-Mgy/R, a = -gy/R

Comparing a=-w²y

w²=g/R

T=2 $π \sqrt[]{\frac{R}{g}}$

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10 months ago

Ranchi University B.Sc.

What is the CUET UG 2025 result date for B.Sc admissions at Ranchi University?

0 Follower 3 Views

Shailja Rawat

Contributor-Level 10

Ranchi University accepts CUET UG scores for admission to its BSc courses. This implies that students willing to take admission can take the CUET UG exam. For the session 2025, the exam was held from 26 May 2025 to 3 Jun 2025. Moreover, the results are expected to be announced in the last week of June 2025.

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10 months ago

Ranchi University B.Sc.

What is the fee structure of Ranchi University for BSc programme?

0 Follower 4 Views

Shailja Rawat

Contributor-Level 10

The fee structure of Ranchi University BSc courses comprises various components such as tuition fees, library fees, exam fees, admission fees, hostel fees, etc. Out of all the components, the total tuition fees range from INR 3,200 to INR 2.2 lakh. Students who wish to know the complete breakdown are advised to visit the official website.

Note: This fee has been sourced from official sources. However, it is subject to change. Hence, it is indicative.

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10 months ago

Patna University Colleges in Patna

What to do after Patna University Entrance Exam result?

0 Follower 2 Views

Tanisha Jain

Contributor-Level 7

The Patna University Entrance Exam 2025 results has been announced online on the official website. The candidate's scores and rank will be mentioned. According to the rank secured, qualified students will be called for counselling to the university. Candidates needs to keep their valid credentials ready to be verified by the authorities while claiming a seat via counselling rounds. Students who make the merit list will be called for the counselling session which will be conducted in the university. There will be different phases of counselling, depending on the seats left and the cut-offs.

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10 months ago

btech Mody University of Science and Technology Courses

Is it worth pursuing BTech from Mody University of Science and Technology?

0 Follower 11 Views

Abhishek Dhyani

Contributor-Level 9

Yes, it is worth pursuing BTech at Mody University of Science and Technology, as it is approved by BCI and AICTE. The university has also been recognised by the UGC. Mody University of Science and Technology, through seven schools of study, offers PhD, UG, and PG courses to students. These courses are provided across the Engineering, Management, and various other streams. The university offers a BTech programme to students at the UG level across several specialisations such as Computer Science and Engineering, Electrical Engineering, Mechanical Engineering, Information Technology, etc.

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10 months ago

Exam MBSE HSLC

Is it necessary to pass the model exam to get the MBSE HSLC hall ticket?

0 Follower 6 Views

Salviya Antony

Contributor-Level 10

No, it is not mandatory to pass the model exam to get the MBSE HSLC hall ticket. Model exams are meant for students to be familiar with the board exam. Clearing the model exams is not a criteria for obtaining the MBSE 10th hall ticket.

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10 months ago

MHTCET

I have 79.49 percentile in MHT CET. Which college and branch can I get?

1 Follower 4 Views

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10 months ago

Exam MBSE HSLC

Is the MBSE HSLC result released online 100% correct?

0 Follower 1 View

Salviya Antony

Contributor-Level 10

Students need not worry about the accuracy of the MBSE HSLC result released online. MBSE 10th result 2026 will be released after various checking process. There is very less chance for the MBSE HSLC result released online to be incorrect. Students are advised to collect the mark list from the school after the MBSE HSLC result is announced.

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10 months ago

physics ncert solutions class 11th Oscillations

One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of 45° each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Let us consider an infinitesimal liquid column of length dx at a height x from horizontal line.

If $ρ =$ density of the liquid

PE= dmgx=A $ρ d x g x$

So total PE of the column

= $\int_{0}^{h_{1}} A d x g ρ x$ = $A g ρ \int_{o}^{h_{1}} x d x = A ρ g \frac{h_{1}^{2}}{2}$

But h₁=lsin45

PE=A $ρ$ gl²sin²45/2

Similarly PE of right column = A $ρ$ gl²sin²45/2

Total PE = A $ρ$ gl²sin²45/2+ A $ρ$ gl²sin²45/2

= A $ρ$ gl²/2

If due to pressure difference is created y element of left side moves on the right side then liquid present in the left arm =l-y

But liquid present in the right arm =l+y

Total PE = PE_final-PE_initial

Change in PE = $\frac{A ρ g}{2} [{(l - y)}^{2} + {(l + y)}^{2} - l^{2}$ ]

= $\frac{A ρ g}{2} [2 {(l}^{2} + y^{2})]$ =A $ρ g {(l}^{2} + y^{2})$

Change in KE = ½ mv²

m=A $ρ 2 l$

change in KE= 1/2A $ρ 2 l v^{2}$ =A $ρ l v^{2}$

...more

This is a long answer type question as classified in NCERT Exemplar

Let us consider an infinitesimal liquid column of length dx at a height x from horizontal line.

If $ρ =$ density of the liquid

PE= dmgx=A $ρ d x g x$

So total PE of the column

= $\int_{0}^{h_{1}} A d x g ρ x$ = $A g ρ \int_{o}^{h_{1}} x d x = A ρ g \frac{h_{1}^{2}}{2}$

But h₁=lsin45

PE=A $ρ$ gl²sin²45/2

Similarly PE of right column = A $ρ$ gl²sin²45/2

Total PE = A $ρ$ gl²sin²45/2+ A $ρ$ gl²sin²45/2

= A $ρ$ gl²/2

If due to pressure difference is created y element of left side moves on the right side then liquid present in the left arm =l-y

But liquid present in the right arm =l+y

Total PE = PE_final-PE_initial

Change in PE = $\frac{A ρ g}{2} [{(l - y)}^{2} + {(l + y)}^{2} - l^{2}$ ]

= $\frac{A ρ g}{2} [2 {(l}^{2} + y^{2})]$ =A $ρ g {(l}^{2} + y^{2})$

Change in KE = ½ mv²

m=A $ρ 2 l$

change in KE= 1/2A $ρ 2 l v^{2}$ =A $ρ l v^{2}$

so from above eqn

change in PE +change in KE = A $ρ g {(l}^{2} + y^{2})$ +A $ρ l v^{2}$

system being conservative

so change in total energy =0

A $ρ g {(l}^{2} + y^{2}) + A ρ l v^{2}$ =0

Differentiating both with respect to time t

A $ρ g (2 y v) + A ρ l (2 v) a = 0$ =0

Dy/dt =v and dv/dt=a

(gy+la)2A $ρ v$ =0

2A $ρ v = c o n s t a n t$ and 2A $ρ v \neq 0$

La+gy=0

A+(gy/l)=0

d²y/dt²+gy/l=0

d²y/dt²+w²y=0

w= $\sqrt[]{\frac{g}{l}}$

T=2 $π \sqrt[]{\frac{l}{g}}$

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10 months ago

Lloyd Institute of Management and Technology Medicine & Health Sciences

What clubs, societies, and events are active in Lloyd Institute of Management and Technology at Pharmacy campus?

0 Follower 4 Views

Nishtha Gupta

Contributor-Level 8

Lloyd Institute of Management and Technology (Pharm.) conducts events such as Niyukti which is Lloyd's flagship event. Apart from this, the Institute conducts various academia centric events that provides industry exposure. The Institute recently conducted various events such as Health-A-Thon 2.0, industrial visit to Yakult Danone India Pvt. Ltd., guest lecture by Multani Pharmaceuticals and many more

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10 months ago

L M College of Pharmacy Pharmacy

I have 99.75% in HSC. Can I get admission for B.Pharma at L M College of Pharmacy?

0 Follower 4 Views

New Question

10 months ago

physics ncert solutions class 11th Oscillations

A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period. 2 m T A g π ρ = where m is mass of the body and ρ is density of the liquid.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Let the log be passed and the vertical displacement at the vertical displacement at the equilibrium position

So mg= buoyant forces = $ρ A x_{o} g$

When it is displaced by further displacement x, the buoyant force is A (x_o+x) $ρ g$

Net restoring force = buoyant forces -weight

=A (x_o+x) $ρ g$ -mg

=A $ρ g x$

As displacement x is downward and restoring force is upward

F_restoring₌-A $ρ g x$ =-kx

So motion is SHM

Acceleration a=F_restoring/m=-kx/m

a=-w²x

w²=k/m

w= $\sqrt[]{\frac{k}{m}}$

T= 2 $π \sqrt[]{\frac{m}{k}} = 2 π \sqrt[]{\frac{m}{A ρ g}}$

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10 months ago

Ganga Institute of Technology and Management

How are the faculty members at Ganga Institute of Technology and Management?

0 Follower 3 Views

Upasana Kumari

Contributor-Level 10

Ganga Institute of Technology and Management appoints well-trained and expert faculty members with years of experience in the related field. The faculty focuses on theoretical and practical knowledge for the students. The faculty engages the students with the latest activities and innovative teaching, and quality research. The faculty focuses on therotical and practical knowledge of the students.

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10 months ago

Humanities & Social Sciences B.A

What is the fee structure for the BA course at North East Frontier Technical University?

0 Follower 2 Views

Himanshu Singh

Contributor-Level 10

The tuition fee for the BA course at North East Frontier Technical University is INR 96,000. This is part of a broader fee structure that may include application and hostel fees. The mentioned amount is based on official sources and is subject to change.

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10 months ago

Exam MBSE HSLC

What if I am not satisfied with my MBSE HSLC result?

0 Follower 1 View

Salviya Antony

Contributor-Level 10

Those who are not satisfied with their MBSE HSLC may opt for re-evaluation. The re-evaluation process for the MBSE HSLC will commence soon after declaration of result. The candidates opting for re-evaluation will have to fill-up the required application form and the prescribed fees. MBSE HSLC Revaluation and Scrutiny results will be out in June 2026.

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10 months ago

Ganga Institute of Technology and Management

How is the infrastructure at Ganga Institute of Technology and Management?

0 Follower 3 Views

Liyansha Katariya

Contributor-Level 10

Ganga Institute of Technology and Management has a state-of-the-art infrastructure which is equipped with all the necessary amenities. The facilities include smart classrooms, a library, a bank, a gym, transportation, a cafeteria, and others. The institute has a well-equipped hostel with all the necessary amenities for the students.

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10 months ago

Mody University of Science and Technology

Mody University of Science and Technology a private or government college?

0 Follower 3 Views

Manori Datta

Contributor-Level 9

Mody University of Science and Technology is a private college and was established in 2004. It is located in Sikar, Rajasthan, and has been accredited by NAAC with an A+ Grade. Mody University of Science and Technology, Lakshmangarh, Sikar (Rajasthan) has been recognised and also certified as a Scientific & Industrial Research Organization (SIRO) by the Department of Scientific and Industrial Research (DSIR).

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